The substance whose Lewis structure shows three covalent bonds is Nitrogen gas molecule.
<h3>What is Lewis structure?</h3>
Lewis structure is a dot structure which gives idea about the number of valence electrons that are involved in the bonding within the molecule.
Lewis dot structure of nitrogen gas will be expressed as in the attached image, where between two nitrogen atoms triple bond is present. That triple bond is formed by the sharing of electrons and known as covalent bonds.
Hence in nitrogen gas molecule three covalent bonds is present.
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Answer:
6.24%
Explanation:
Molality by definition means a measurement of the number of moles of solute in solution with 1000 gm or 1Kg solvent. Notice the difference that Molarity is defined on the volume of solution and Molality on the mass of solvent.
So, An aqueous solution of iron(II) iodide has a concentration of 0.215 molal.
means 0.215 moles are present in 1 Kg of solvent.
The molar mass of Fe2I = 309.65 g / mole
mass of FeI2 = moles x molar mass
= 0.215 x 309.65
=66.57 gm
mass % of FeI2 = mass of FeI2 x 100 / total mass
= 66.57x 100 / (1000 +66.57)
= 6.24%
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
